unit 2 vectors 2 - ignou

28

Vectors and Three

Dimensional Geometry UNIT 2 VECTORS – 2

Structure

2.0 Introduction

2.1 Objectives

2.2 Scalar Product of Vectors

2.3 Vector Product (or Cross Product) of two Vectors

2.4 Triple Product of Vectors

2.5 Answers to Check Your Progress

2.6 Summary

2.0 INTRODUCTION

In the previous unit, we discussed vectors and scalars. We learnt how to add and

subtract two vectors, and how to multiply a vector by a scalar. In this unit, we

shall discuss multiplication of vectors. There are two ways of defining product

of vectors. We can multiply two vectors to get a scalar or a vector. The former

is called scalar product or dot product of vectors and the latter is called vector

product or cross product of vectors. We shall learn many applications of dot

product and cross product of vectors. We shall use dot product to find angle

between two vectors. Two vectors are perpendicular if their dot product is zero.

Dot product helps in finding projection of a vector onto another vector. The cross

product of two vectors is a vector perpendicular to both the vectors.

If cross product of two vectors is zero then the two vectors are parallel (or

collinear). Cross product of vectors is also used in finding area of a triangle or a

parallelogram. Using the two kinds of products, we can also find product of

three vectors. Many of these products will not be defined. In this unit, we shall

discuss the two valid triple products, namely, the scalar triple product and the

vector triple product.

2.1 OBJECTIVES

After studying this unit, you should be able to:

define scalar product or dot product of vectors;

find angle between two vectors;

find projection of a vector on another vector;

29

Vectors - II define cross product or vector product of vectors;

use cross product to find area of a parallelogram vector product of vectors;

define scalar triple product and vector triple product of vectors.

2.2 SCALAR PRODUCT OF TWO VECTORS

Definition : The scalar product or the dot product of two vectors and ,

denoted by . is defined by

. = | | | |

where is the angle between the vectors and .

Also the scalar product of any vector with the zero vector is, by definition, the

scalar zero. It is clear from the definition that the dot product . is a scalar

quantity.

Sign of the Scalar Product

If and are two non zero vectors, then the scalar product

. = | | | | cos

is positive, negative or zero, according as the angle , between the vectors is

acute, obtuse or right. In fact,

is acute cos 0 . 0a b

is right cos 0 . 0a b

is obtuse cos 0 . 0a b

Also, note that if and and are non zero vectors then . = 0 if and only if

and are perpendicular (or orthogonal) to each other.

If . is any vector, then the dot product . ., of with itself, is given by

. = | || | cos 0= | |2

Thus, the length | | of any vector is the non negative square root , i.e., of

the scalar product , i.e.,

| | = ,

Angle between two Vectors

If is an angle between two non zero vectors and , then

. = | || | cos

30

Vectors and Three

Dimensional Geometry

So, the angle between two vectors is given by

Properties of Scalar Product

1. Scalar product is cumulative, i.e., . = . for every pair of vectors and .

2. . (– ) = – ( . ) and (– ). (– ) = . for every pair of vectors and .

3. ( λ1 ) . (λ2 ) =( λ1 λ2) ( . ) where and are vectors and λ1 , λ2 are scalars

4. (Distributivity) .( + ) = . + .

The following identities can be easily proved using above properties.

(i) ( + ) . ( – ) = a2 – b

2 (here, a

2 = . = | |

2)

(ii) ( + )2

= a2 + 2 . + b

2

(iii) ( – )

2 = a

2 – 2 . + b

2

If , and are mutually perpendicular unit vectors, then

. = . = . = 1 and

. .= . = . = 0 (1)

If and be two vectors in component

form, then their scalar product is given by

1 1 2 3 2 1 2 3 3 1 2 3

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ . ( + ) ( + ) ( + ) (using distributivity)a i bi b j b k a i bi b j b k a k bi b j b k

2 3 3 1 3 2 3 3ˆ ˆ ˆ ˆ ˆˆ= ( . ) ( . ) ( . ) ( . ) (using properties)a b j k a b k i a b k j a b k k

= a1 b1 + a2 b2 + a3 b3 (using (1))

Thus . = a1 b1 + a2 b2 + a3 b3 .

Projection of a Vectors on another Vector

Let = be two vectors

A

Figure 1

O L B

31

Vectors - II Drop a perpendicular form A on OB as shown in Figure 1. The projection of

on is the vector . If is the angle between and , then projection of

and has length | cos and direction along unit vector . Thus, projection

of on = (| | cos )

2

. .

| | | |

a b a bb b

b b

The scalar component of project of on is Similarly, the scalar

component of projection of on

a . ba b .

|b |

a . bb a

| a |

Example 1: Find the angle between the following pair of vectors.

(a) +

(b)

Solution: (a) . = (2 + 3 ) . ( + 4 )

= 2.1 + 0.4 + 3 ( 1)

= 2 3 = 1

= . = (2 + 3 ) . (2 + 3 )

= 2.2 + 0.0 + 3.3

= 13

|a| =

= . = ( +4 – ) . ( +4 )

= 1.1 + 4.4 + ( 3) ( 3)

= 26

|a| =

(b) Here, . = ( + 2 + ). (2 + 2 + )

= 1.2 + 2.(–2) + 2.1

= 0

= = 3

32

Vectors and Three

Dimensional Geometry = = 3

.

Example 2 : Show that is perpendicular to – , for any

two non zero vectors .

Solution : We know that two vectors are perpendicular if their scalar product is

zero.

( ) . ( )

= ( ) + . ( )

(using distributivity)

( ) ( ) + ( ) ( )

( ) + ( ) = 0

so, the given vectors are perpendicular.

Example 3 : Find the scalar component of projection of the vector

ˆ ˆˆ ˆ ˆ 2 3 5 on the vector 2 2 . a i j k b i j k

Here,

2 22and |b| = = 3( 2) ( 1)2

Example 4 : Show that the diagonals of a rhombus are at right angles.

Solution : Let A B C D be a rhombus (Figure 2 )

D C

A B (Figure 2)

33

Vectors - II ABCD being a rhombus, we have

=

= ( + ) . ( )

=

= – = 0

2 2

( )AB AD

Example 5 : For any vectors and , prove the triangle inequality

| + | ≤ | | + | |

Solution : If = or = , then the inequality holds trivially.

So let | | ≠ 0 ≠ | |. Then,

= ( + )2

= ( + ). ( + )

=

= | |2 2 + | |

2 ( = )

= | |2

2| | |cos | |2

where is the angle between and

| |2

2| | | | |2

(cos 1 )

= (| | | |) 2

Hence | | | | | |

Remark : Let = and = , then + =

C

+

A B

Figure 3

As shown in figure 3 inequality says that the sum of two sides of triangle is

greater than the third side. If the equality holds in triangle inequality, i.e.,

| + | | | +| |

Then | AC

| = | AB

| + | BC

|

showing that the points A, B and C are collinear.

Check Your Progress – 1

1. If = 5 –3 –3 and = + –5 then show that + and – are

perpendicular.

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Vectors and Three


2. Find the angle between the vectors

(a) = 3 – +2 = – 4 +2

(b) = – –2 = –2 +2 +4

3. Find the vector projection of on where =3 –5 +2 and = 7 + –2 .

Also find the scalar component of projection of vector on .

4. Prove the Cauchy – Schwarz Inequality

| . | ≤ | | | |

5. Prove that | – |2 = | |

2 + | |

2 – 2 .a b

2.3 VECTOR PRODUCT (OR CROSS PRODUCT) OF TWO

VECTORS

Definition : If and are two non zero and non parallel (or equivalently non

collinear) vectors, then their vector product × is defined as

= | | | |sin

where is the angle between and (0 < < ) and is a unit vector

perpendicular to both and such that , and form a right handed system.

If and are parallel (or collinear) i.e., when = 0 or , then we define the

vector product of and to be the zero vector i.e., = . Also note that if

either or ,then

Properties of the Vector Product

1. = since = 0

2. Vector product is not commutative i.e., ≠ .

However, = –

We have = | || |sin where , and form a right handed system and

= | || | |sin | where , and ight handed system. So the

direction of | is opposite to that of

Hence, = | || |sin

= –| || |sin

= – .

3. Let from a right–handed system of mutually perpendicular unit vectors

in three dimensional space. Then

= = = and

= =

Also, = – = –

a

Figure 4

35

Vectors - II 4. Two non zero vectors and are parallel if and only if

=

5. Vector product is distributors over addition i.e., if , and

are three vectors, then

(i) ( ) = .

(ii) ( ) = .

6. If is a scalar and and are vectors, then

( = (

Vector Product in the component form

Let = a1 a2 a3 and

= b1 b2 b3

Then

= (a1 a2 a3 ) (b1 b2 b3

= a1 b1 ( ) a1 b2 ( ) a1 b3 ( ) a2 b1 ( ) a2 b3

( ) a2 b3 ( ) a3 b1 ) a3 b2 ( ) a3 b3

( )

= a1 b2 a1 b3 (– a2 b1 (– a2 b3 a3 b1 a3 b2 (–

= (a2 b3 a3 b2) (a1 b3 a3 b1) (a1 b2 a2 b1)

=

Example 6: Find if = and = 3

Solution : We have

=

= (3 – 2) – (3 – 1) + (2 – 1)

= – 2

Example 7: Find a unit vector perpendicular to both the vectors

ˆ ˆˆ ˆ ˆ 4 3 and 2 2 a i j k b i j k

Solution : A vector perpendicular to both and is

=

= (2 – 3) – (–8 6) (4 – 2)

= – 2

| | = = = 3.

So the desired unit vector is

36

Vectors and Three


Example 8 : Prove the distributive law

( ) a b c a b a c

using component form of vectors.

Solution : Let = + +

= + + and

= + + .

So, + = + + ( +

Now,

× ( + ) =

=

= + .

Example 9: Show that

| |2

= (

Solution : | |2

= | || | sin

| |2

= (| || | sin )2

= | |2| |

2 (1– cos

2)

= | |2| b

|2 – ( | |

2| |

2 cos

2)

2 2 ( ) ( ) ( . )aa bb a b

Example 10 : Find | | if

| | = 10, | | = 2, = 12.

Solution : Here = 12

a

a

37

Vectors - II Area of Triangle

Let , be two vectors and let be the angle between them ( O < < ). Let O

be the origin for the vectors as shown in figures (Fig. 5) below and let

= , =

B D

O C A Figure 5

Draw BC OA.

Then BC = OB Sin = |b| sin

Thus, if and represent the adjacent sides of a triangle, then its area is given as

Area of a Parallelogram

In above figure, if D is the fourth vertex of the parallelogram formed by O, B, A

then its area is twice the area of the triangle OBA.

Hence, area of a parallelogram with and as adjacent sides = |

Example 11: Find the area of ABC with vertices A (1,3,2), B (2, –1,1) and

C(–1, 2, 3).

Solution : We have

= –

= (2 – 1) (–1 – 3) (1 – 2)

= – 4 –

and = ( –1 – 1) (2 – 3) (3 – 2)

= – 2 –

θ

38

Vectors and Three


Example 12 : Show that ( – ) ( ) = 2( ). Interpret the result

geometrically.

Solution : ( – ) ( )

= ( – ) ( – )

= – –

= 0 0

= 2 ( )

Let ABCD be a parallelogram with = and = – .

Then area of parallelogram = =

D C

b

A a

B Figure 6

Also, diagonal = +

and diagonal = –

( ) ( ) =

= area of parallelogram formed by and .

Thus, the above result shows that the area of a parallelogram formed by diagonals

of a parallelogram is twice the area of the parallelogram.


1. Find a unit vector perpendicular to each of the vector ( + ) and ( – ),

where = 2 – 4 and = – 2

2. Show that

( ) ( ) ( ) =

39

Vectors - II 3. For Vectors

= – 2

Compute ( and ( ).

is ( × = ( ).

4. If the vectors , and satisfy + + = 0, then prove that

= =

5. Find the area of a parallelogram whose diagonals are

= 3 + –2

2.4 TRIPLE PRODUCT OF VECTORS

Product of three vectors may or may not have a meaning. For example,

( . as is a scalar and dot product is defined only for

vectors. Similarly, ( . has no meaning. The products of the type

( and are meaningful and called triple products. The

former is a vector while the latter is a scalar.

Scalar Triple Product

Definition : Let , and be any three vectors. The scalar product of . )

is called scalar triple product of , and and is denoted by [ , ]. Thus

[ , ] = .( .

It is clear from the definition that [ , ]. Is a scalar quantity.

Geometrically, the scalar triple product gives the volume of a parallelepiped

formed by vectors , as adjacent sides.

Scalar Triple Product as a determinant

Let =

and =

Then

.(

= ( –

= –

40

Vectors and Three

Dimensional Geometry =

Thus, .(

Note : We can omit the brackets in .( because

( . is meaningless.

Properties of Scalar Triple Product

1. [ , = [

= – [ ] = – [ ,

This is clear if we note the properties of a determinant as [ , can be

expressed as a determinant.

2. In scalar triple product .( , the dot and cross can be interchanged.

Indeed, .( = [ = [

= =

3. [ , , ] [ , , ]pa qb rc pqr a b c

where p, q and r are scalars. Again it is clear

using properties of determinants.

4. If any two of are the same then [ For example,

[ ,

Coplanarity of three vectors

Theorem : Three vectors are coplanar if and only if [

Proof : First suppose that are coplanar . If are parallel

vectors, then If are not parallel,

then being perpendicular to plane containing is also

perpendicular to because

Conversely, suppose that [ , = 0. If are both non zero, then

are perpendicular as their dot product is zero. But is

perpendicular to both and hence must lie in a plane, i.e.,

they are coplanar. If = then are coplanar as zero vector is

coplanar with any two vectors. If = , then

, must be coplanar.

Note : Four points A, B , C and D are coplanar if the vectors , and are

coplanar.

41

Vectors - II Example 13 : If = 7 – 2 +3 = – 2 +2 , and = 2 + 8

find [ , ,

Solution : [ , , = . ( ×

= 7(0 – 16) + 2(0 – 4) + 3(8 + 4)

= –112 – 8 + 36

= – 84

Example 14 : Find the value of λ for which the vectors

= – 4 + = λ – 2 + , and = 2 + 3 + 3 are

coplanar,

Solution : If , and are coplanar, we have

= . ( = 0 i.e.,

1(–6 –3) + 4(3 – 2) + (3 + 4) = 0

Example 15 : If , , are coplanar then prove that , and are also

coplanar.

Solution : Since , , are coplanar,

[ , ,

Now [ , ] = ( ) .( )

= ( + ) .[( + ( ]

= ( + ) .[ × ] ( as 0

= .( a

.( ) + .( ) + ( + ( ) +

= [ , , [ , , +

= 2 [ , , (using property (4)

= 0

42

Vectors and Three


Vector Triple Product

Definition : Let , , and any three vectors. Then, the vectors ( ) and

( ) are called vector triple products.

It is clear that, in general

( ) ( ) ×

Note that ( ) is a vector perpendicular to both and . And also

is perpendicular to both and . Thus ( ) lies in a plane containing

the vectors and , i.e., ( ), and are coplanar vectors.

We now show that for any three vectors ,

( ) = ( – (

Let =

and = Then

=

= ( – + ( – + ( –

(

– – –

= [ ( – + – (

+ [ (

Also, (1)

= (

(

= [ (

[ (

[ (

= [ (

[ (

[ ( (2)

43

Vectors - II From (1) and (2), we have

( = ( – (

It is also clear from this expression that ( is a vector in the plane of

and

Now = – (

= – [ – ( )

= ( )

= ( )

So, ( is a vector in the plane of .

Theorem : Let be any three vectors. Then

if and only if are collinear.

Proof : First suppose

Now, (

and ( (

, ( . ) ( . )So a b c b c a

Conversely, suppose that

. Then

– ( –

and – ( –

So, – (

( ( =(

i.e., (

Example 16 : Show that

Solution :

= ( – ( . ) – ( ) + ) – (

= since dot product is commutative.

Example 17 : Let be any four vectors. Then prove that

(i) ( ) = [ ] c

[

(ii) ( ) = [ ] [

44

Vectors and Three

Dimensional Geometry Solution : (i) Let = Then

( ) = )

= (

= (

= [ –

= [

= [

(ii) Let = . Then

( ) = (

(

= – [( r

– (

= (

= (

= ( ) – ( )

= [ , ] – [ ]

Example 18 : Prove that

[ , , a

=

Solution : [ , , a

= ). [ × ) (

= )[ ) )

= ). [ ) a

c

] ( )

= ) [ ) ] = [ a

)][ )]

= [ ]2.

Example 19 : For vectors =

and = verify that

) )

Solution : We have

× =

= 3 + 3 + 3

45

Vectors - II So, (

=

= – 9 –6 – 3 (1)

Also,

= ( –2 + .( + 2 +

= 1 – 4 – 1 = – 4

( ) = – 4 ( + + = – 8 – 4 – 4

and = ( +2 + .( + + = 1

( ) = + 2 –

Thus, ( ) – ( ) = – 6 – (2)

Hence, from (1) and (2)

) = ( ) – ( )


1. For vectors = + , = +2 = + +2 find [ , ].

2. Find the volume of the parallelepiped whose edges are represented by

= – 3 , = + = – +2

3. Show that the four points having position vectors

+ , 6 +11 +2 + 6 i + + 4 are coplanar.

4. Prove that

( ). ( ) = ( . )( . ) – ( )( . )

5. For any vector prove that

( ) + ( ) + ( ) = 2

6. Prove that

[ ( )] = ( . ) ( × )

2.5 ANSWERS TO CHECK YOUR PROGRESS


1. Here, + = + 2 and

– = – 4 + 2

So, ( + ). ( – ) = ( + 2 – 4 + 2

= 24 – 8 – 16 = 0

Hence + and + are perpendicular vectors.

47

Vectors - II Check Your Progress – 2

1. Here = 2 +

= 3

Let = . Then is vector perpendicular to both

Now =

= (–6 + 6)

= 12

A unit vector in the direction of is

So, is a unit vector perpendicular to both

2. ) a

( )

= (using distributivity)

= – – (

= – and = – = 0

3. = –

= – 5

Also,

=

= –

Clearly,

(

48

Vectors and Three

Dimensional Geometry 4.

i.e., +

i.e., + 0

(

i.e., = –

i.e., = ……………..(1)

Similarly,

=

i.e., +

i.e., = –

i.e., = ……………. (2)

From (1) and (2), we have

=

5. Let ABCD be the parallelogram

D C

A B

Figure 7

=

So area of parallelogram ABCD

= |

50

Vectors and Three


, and are coplanar.AB BC CD

Hence A, B, C and D are coplanar.

4. Let × = Then

( ). ( ) = r

.( )

= ( r

).

= (( ) ).

= [(

= ( ) – ( . )

= ( ) – ( )

5. L.H.S. = ( ) ( ) ( )

= [ ( ) i [ ( )

= )

= i )

Let =

So, –

Similarly, =

L.H.S. = 3 –

= 3

6.

= .

= (

= – ( ( 0

= ( (

51

Vectors - II 2.6 SUMMARY

This unit discusses various operations on vectors. The binary operation of scalar

product is discussed in section 2.2. In the next section, the binary operation of

vector product (also, called cross product) is illustrated. Finally in section 2.4, the

ternary operation of tripe product of vectors is explained.

Answers/Solutions to questions/problems/exercises given in various sections of

the unit are available in section 2.5.

unit 2 vectors 2 - ignou

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